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MOTION OF A PARTICLE
123
In actual practice the angle of elevation which gives the maximum range is smaller on account of the resistance of the air.
THE HIGHEST POINT.—At the highest point y = 0.    Therefore substituting this value of y in equation (4) we obtain
-------- or - T for the time taken to reach the highest point.
9            ^
Substituting this value of the time in equation (8) we get for the maximum elevation
„ _ flQ2 Sm2 a.                                         ,
—2~g—"                             '    '
THE RANGE FOE A SLOPING GROUND. — Let  0 be  the
angle which the ground makes with the horizon.    Then the
range is the distance OP, Fig. 67, where P is  the point where the projectile strikes the sloping ground. The equation of the line OP is
y = x tan 0.                               (14) Y
\
xp                           X.
FIG. 67.
Eliminating y between equations (14) and (9) we obtain the re-coordinate of the point,
~  - 2 VQ* cos2 a (tan a. — tan 0)
But Therefore
R' cos ft   where   R' = OP.
COS PL _.      f
r— Sin (a — 2
g cos2 j8 flp2 sin (2 ex — ft)-- sin
cos2
(15)